Question

The normal to the curve $y(x-2)(x-3)=x+6$ at the point where the curve intersects the y-axis passes through the point.

• A. $\left(-\frac{1}{2},-\frac{1}{2}\right)$
• B. $\left(\frac{1}{2},\frac{1}{2}\right)$
• C. $\left(\frac{1}{2},-\frac{1}{3}\right)$
• D. $\left(\frac{1}{2},\frac{1}{3}\right)$

Answer 1

Answer: (B)

$\left(\frac{1}{2},\frac{1}{2}\right)$

Given $y(x-2)(x-3)=x+6$

$⟹y=\frac{x+6}{(x-2)(x-3)}$

Taking derivative on both the sides w.r.t $x$ we get

$⟹\frac{dy}{dx}=\frac{(x^2-5x+6)\left(1\right)-(x+6)(2x-5)}{(x^2-5x+6{)}^2}$

$⟹\frac{dy}{dx}=\frac{-x^2-12x+36}{(x-2{)}^2(x-3{)}^2}$

At $x=0$ we get $y=1$

$⟹\frac{dy}{dx}=1$ at $x=0,y=1$

Tangent at $x=0,y=1$ is $\frac{dy}{dx}=1$

Slope $=-\frac{1}{tangent}=-1$

Normal is given by

$y-1=(-1)x$

$y-1=-x$ ......... $\left(1\right)$

Now, at $(\frac{1}{2},\frac{1}{2})$

$y-1=\frac{1}{2}-1=-\frac{1}{2}=-x$

$\therefore$ Equation $\left(1\right)$ satisfies at $(\frac{1}{2},\frac{1}{2})$

Hence, normal passes through the point $(\frac{1}{2},\frac{1}{2})$.